Design of Systems - Force Flo of Systems 2009 Training Program 3 Objectives of Water Treatment •...

Design of Systems 2009 Training Program

1

Design of Fluoridation Facilities

You have heard about the fluoride Additives and equipment , now we will try to put them together in the design process.We will try to discuss the considerations on design of fluoridation

systems. Since this class has a wide range of participants extending from program specialists who may not have a technical background, to practicing engineers, it is difficult to have a course that fully addresses everyone’s knowledge level of needs. But we are going to try to include enough background information for the non-engineers so that you understand many of the issues involved, but also provide enough technical content so the engineers here have a basis for looking at the designs.


2

Water Utility Configuration

Elements of a water utility• Source: well, river, lake, or purchase

• Treatment: excellent quality eligible to waive

• Distribution: pipe network

• Customer service: (billing, maintenance, complaints/problems, management)

A water system infrastructure includes four essential elements.

The first thing to look at is the source of the water. Is it a well water, a surface river, a lake or impoundment, or does the watersystem purchase their water? All of these can influence the considerations on water fluoridation. As an example, a well water of excellent quality may not require any treatment, so adding fluoride might be more difficult for such a system not used to processing the water.

The treatment facility, or facilities, are another consideration. Are the facilities old or new. Is there a single facility or multiple facilities. What type of staffing is required?

The third element of a water system is the distribution pipe network which delivers the water to the customers.


3

Objectives of Water Treatment

• Disinfection to protect against pathogenic organisms

• Reduction of contaminants for health effects

• Aesthetics – Taste, Odor, and color

The objectives of water treatment are •Disinfection, for pathogenic organisms•Reduction of contaminants for health effects•Oxidation of undesirable reduced compounds•Improvement of the aesthetics: taste, odor, and color


4

Simplified Water Treatment

Pretreatment Turbidity Removal Disinfection Effluent Finishing

Water Softening Disinfection Effluent Finishing

OtherContaminant

Reduction

Fluoride Addition

Here is a simplified schematic diagram of water treatment. The top line would be a conventional surface water treatment facility flow train, and the bottom one would be groundwater system that requires some processing of the extracted water. You will see that water fluoridation would be like corrosion control in that it would be done as part of effluent finishing as the water is delivered to the water system for distribution.


5

Water Treatment - Disinfection

• Objective: to protect against pathogenic organisms

• Methods:

−hypochlorous acid (HClO), Hypochlorite (ClO-)

−Sodium hypochlorite, calcium hypochlorite

−Chloramines

−UV (ultra-violet irradiation)

−Ozone


6

Water Treatment - Turbidity

• Objective: to protect against enteric vectors and pathogenic organisms

• Methods:

−Coagulation/flocculation/sedimentation

• Alum (aluminum sulfate)

• Iron (ferric, ferrous compounds)

• Polymers

−Filtration (normally rapid sand)


7

Water Treatment - Softening

• Objective: to reduce water hardness (divalent ions) to avoid soap interference and scaling from Carbonate/non-carbonate hardness (bicarbonates) and permanent hardness (remaining scale upon boiling)

• Methods:

−Lime-soda

−Zeolite base-exchange resin


8

Water Treatment - Aesthetics

• Objective: to improve characteristics for consumer satisfaction

• Methods:

−Aeration – volatile release such as H2S

−Oxidation – manganese/iron, etc, using potassium permanganate, chlorine, chlorine dioxide

−Nuisance – color, algae, etc


9

Water Treatment - Adsorption

• Objective: to reduce organics and hydrocarbons

• Methods:

−Powdered activated carbon

−Granular activated carbon


10

Water Treatment - Membranes

• Emerging technologies replacing traditional treatment processes

• Methods:

−Ultra-filtration – low pressure to remove particulate and colloidal contaminants

−Reverse osmosis – high pressure to remove ions


11

Water Treatment - Stability

• Objective: to decrease corrosive nature of water

• Methods:

−Lime (carbonates)

− (Ortho- or poly-) phosphates

−Silicates


12

Water Treatment - Fluoridation

• Objective: to promote community oral health

• Method

−Convince community they should do the right thing even though the EPA or State drinking water program does not say they must do it

Water fluoridation requires a community to decide that it is theright thing to do, not because it is a requirement.


13

Before You Begin…• Ensure project scope understood

− Features− Capacity− Previous reviews of facility (sanitary surveys,

deficiencies, compliance problems)− Space requirements− Building codes− Fire Marshal requirements

• Types of facilities?− Storage− Feed equipment− Sampling

• Visit other communities to learn their successes and failures

Before you begin, it is important that all parties clearly understand the project. Think about what is important, what is good, what would be bad, and issues to be addressed during design. Check on Fire Marshall requirements for handling hazardous wastes in the jurisdiction, which building codes will be used for issuing a construction permit, and ask the drinking water program if the facility had a previous facility review such as a sanitary survey, deficiencies, or compliance problems of which you should be aware. What types of facilities will be necessary? And you can often learn a lot by visiting other communities to learn about their successes and failures.


14

Permitting

• State construction permits− Expansion of environmental facilities− Erosion and drainage− Spill Prevention and Control Plans

• Air quality permits –− California− Metropolitan non-attainment areas

• Building Permits− Responsible entity− Local rules and requirements− Special requirements

• Neighborhood approvals

Permitting can be a very expensive and time consuming aspect of a project.

State permits could include permits to allow expansion or construction of improvements to drinking water facilities, and many locations now require erosion and drainage plans. Also, ifyou are using FSA and facility exceeds the thresholds for tank capacities, you may need to comply with SPCC requirements.

Air quality may be necessary if you are in California or in a metropolitan non-attainment area. Normally these have been avoided in the past, but you need to confirm that requirements have not changed.

Building permits have been known to slow or stop a project. Find out which government entity will issue a building permit (City, County, etc) and what their requirements might be (Standard Building Code or other). Check on if there are local rules or special requirements. What does the Plumbing Code require, and is the International Fire Code applicable (500 gallons storage without sprinklers and 1,000 gallons with storage)

Neighborhood permits can sometimes also be necessary if you are in larger urban areas with neighborhood or planning group oversight.


15

Engineering Submittals and Approvals

• State submittal and review often required

• Drinking water administrator and Oral Health program should collaborate to prepare checklist for projects on state requirements

Improvements to water treatment facilities typically require state drinking water program review and approval. Check with the requirements of the state drinking water program on your state’s requirements.

It is advisable that the oral health program and the drinking water administrator collaborate on preparing checklists and guidance on state requirements.


16

If Noah was asked to build an Ark today…

• Local building permit only allows structures conforming to zoning ordinance

• Soil Conservation requires drainage study• Fire Marshall requires wood ark to have sprinklers

even though it will be during a rainstorm• DOT utility encroachment permits bogged done for

clearance issues• Wood availability poor due to timbering restrictions

for spotted owl conservation requirements• Legal injunction by animal rights group protesting

collection of pairs of animals• EPA requires EIS for flood impacts• Immigration and Naturalization reviews of green

card status of employees• Trade union picketing of open labor policies• IRS seizing assets to prevent illegally leaving

country without declaring assets


17

Fluoridation Design Basis

• Basic design principles for fluoridation same as for other water treatment processes

• Equipment and process design same as for other standard water treatment processes

• Check with drinking water group to see if your state has preferences

−Regulatory requirements/checklists

−10-States or other guidance

The design principles for water fluoridation is the same as for other water treatment processes. The calculations are same as for other additives and the equipment and process design would be the same as for other standard water treatment processes.


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It’s really very simple

…You need a tank to hold the FSA or solution mixed from the dry additive, and a pump…

…but the devil is in the details….

It’s really very simple, you need a tank to hold the solution and apump, but as with most things, there are certain details to get it right.


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Standard Reference Guidance

American Water Works Association Manual of Practice (MOP) No. 4 –Water Fluoridation Principles and Practices (fifth edition 2004)

Great Lakes-Upper Mississippi River Board of State and Provincial Public Health and Environmental ManagersRecommended Standards for Water Works (AKA 10 States Standards)

The standard reference guide is the AWWA Manual of Practice No. 4, Water Fluoridation principles and Practices. The currentversion is the Fifth Edition which was published in 2004. If you are adding fluoride to your system, you should purchase this manual of practice from AWWA as a reference. It contains a wealth of information on the practices and operations related towater fluoridation.

The other industry reference is 10-State Standards.


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10-States Standards

• Most common reference by state programs

• Recommendations in this program are consistent with 10-States

• Some highlights−Section 4.7 is Fluoridation−Section 5 is Chemical feed systems−Pump stroke not less than 20 per minute and

not less than 20% of pump capacity−Feeders accurate to 5% of feed rate−Scales, liquid level indicators, etc, accurate to

5% of daily change− Injected at positive pressure or air gap

provided

10-States is the most common reference used by state drinking water programs when reviewing designs. The materials in this CDC training are consistent with 10-States. Although Section 5 on chemical feed systems apply to fluoridation, there is also a Section 4.7 that is specific to fluoridation, so both sections must be checked. There are also certain specific requirements and a few of them include minimum pump strokes, minimum pump capacity, accuracy of measurement, and injection of solutions atpositive pressure locations or with an air gap.


21

Fluoride System Selection

• There is no one specific type of system or equipment that is best

• Historically−Large city - FSA

−Medium system – FSA, dry feeder

−Saturator or dry feeder

There is no one specific type of equipment that is used solely for one type of water system. All the fluoridation equipment, with the exception of a saturator, can be used on any system.

Historically, large water systems typically use FSA, a medium-sized system can use either FSA or dry additives with a feeder, and a very small system uses dry additives with either a saturator or a feeder.

FSA systems are more frequently used because many utilities believe that because the acid is contained in tanks and pipes, there is less hazardous exposure to the operator and require less operator involvement and time. Larger systems typically have found FSA more cost-effective than dry additives in the larger delivery capacities. Some locations with hard water have found that usingFSA avoids scaling and plugging from fluoride precipitation products.

Some utilities have chosen the dry additives because they have fewer impurities than FSA, or because they believe that exposure of personnel to dry additives and the resulting solutions present less hazardous conditions than FSA. So dry additives and FSA advocates can both argue that there is less resulting exposure, depending on assumptions, and they can both be right.

In fact, there are reasons to select any of the systems or additives, and a community should consider all the factors before deciding which system to use.


22

Size does matter

1,000

10,000

100,000

1,000,000

Satu

rato

r Dry

feed

er

Flu

oro

sili

cic

aci

d

Less likely

Less likely

We have seen saturators used for large facilities where the saturator is used in a continuous flow basis to feed a day tank,and we have seen acid feed systems used in extremely small systems serving less than 100 people, but these are exceptions. Very small systems are typically going be least expensive as a saturator system, and very large systems will be least expensiveas an acid feed. If a community desires to use a dry additive, then a dry feeder will become increasingly more cost effective as the size of the population served increases due to reduced laborcosts.


23

How Many Fluoridation Systems?

• Surface water system may have one treatment plant at intake with one fluoride addition location

• Groundwater system may have distributed feed with multiple wells each requiring separate feed system, or a well field with a single point of input

The number of fluoridation systems to serve a community depends on several factors. If the water system has a surface water supply, like a lake or a river, then there may be only onetreatment facility, requiring a single fluoride addition location. If the community has two or more treatment facilities, then each location requires a separate feed system.

If the community is served with a groundwater system, then theremay be multiple wells, resulting in a distributed feed to the distribution system. If that is the case, then each well requires a separate feed system.

Two equally sized communities, but one with a surface water plant and the other with a distributed well system, might have completely different fluoridation systems. The single surface water plant might use an FSA feed system, whereas the well system might be better off with a saturator or volumetric feeder for dry additive at each wellhead.


24

System complexity considerations

Water System

Surface water


25

Fluoride Additive Selection

• Fluoride Products Availability

• Existing Facilities−Compatibility with water system

−Availability of space

−Number of treatment sites (fluoride injection points)

• Characteristics of the Water−Natural fluoride level

−Type of flow (variable or steady state)

−Pressure (discharge)

How do you design a water fluoridation system?

First, you select the additive and then select the equipment. The compatibility with the water system and other equipment may be a factor, such as using the same brand of metering pump or volumetric feeder that other chemical additive locations use in the facility. The type of flow, such as variable feed or steady state, and the pressure at the discharge point (gravity discharge into a tank or flume versus injection to a pipe) are other considerations. The number of treatment sites can also figure into the decision.

The natural fluoride level must be checked along with the optimum level for addition of fluoride.


26

Each Installation is Unique

Each installation will require consideration of which additive, how much, how delivered, how the space is configured, pacing and controls, and client preference

Engineers earn their keep by considering the unique elements of each specific installation.


27

Fluoride Additive Selection (continued)

• Estimated overall cost −Capital (initial)

−Operation and Maintenance (O&M)

−Chemicals

• State rules, regulations, and preferences

Cost is a major—and, in many cases, the deciding—factor in the design of a fluoridation system, especially capital costs.

Of course, state rules and regulations have to be followed and may affect the design a great deal. Sometimes there is a preference by state engineers for a certain additive or installation, so check with the drinking water program for the state for theiropinion.


28

Process Calculations - Units

−MGD, gpm, m3/day

• Fluoride dosage−mg/L = ppm

• Fluoride additive feed rate− lb/day, lb/hr, mL/min, gpd, gph

• Capacity

To get the right dosage of fluoride additive, it is necessary toknow how much to add. This requires some math, but the calculations are the same type as those required for other chemical additives at a treatment facility. So let’s review the simple math calculations that are needed for any process controlin a water facility. As is the case with other calculations, it is normally necessary to express the calculations in units of measurement.

For the plant flow, million gallons per day (MGD), or gallons per day (gpd) for smaller facilities, and cubic meters per day (m3/day) in the metric system.

For the dosage, either milligrams per liter (mg/L) or parts per million (ppm), which is approximately the same thing.

The additive feed rate is normally expressed as pounds per day (lb/day) or pounds per hour (lb/h) for dry additives, or as milliliters per minute (mL/min), gallons per day (gpd), or gallons per hour (gph) for FSA feed.


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Process Calculations

−Amount of fluoride chemical needed to obtain optimal level

• Optimal level

• Natural level

• Desired dosage

level Natural - level Optimal Dosage =(mg/l) (mg/l) (mg/l)

To determine the amount of fluoride additive, you need to know the optimal level for your location and what the natural fluoride level is for your source water. The optimal level can be determined by consulting with your state water fluoridation program, and the natural level must be measured. Some locations will find the natural level to be stable, but other locations will find it changes based on seasonal influences. Once you knowthe target optimal level and the natural level, you can determine the necessary dosage by simple subtraction. For example, if the optimal level is 0.8 mg/L, and the natural level is 0.2 mg/L, then the dosage is 0.6 mg/L.


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− Maximum hour capacity (2.0)− Maximum day capacity (1.4)− Maximum month capacity (1.0)− Annual average capacity (0.8)− Actual capacity (could be any of the rates)− Actual operating rate (measured)− Minimum capacity (0.2)

• Groundwater well delivers constant volume, often to a storage tank

• Treatment facility capacity

Once the dosage has been calculated, then the quantity of fluoride additive can be calculated. This requires knowledge of the treatment facility capacity. Here is a listing of various capacity measurements for one treatment facility. Typically, a water facility is rated to have the capacity to produce the maximum month demand, and that is often the measurement we think of as a facility capacity. But, in fact, there are other measures of capacity including maximum hour, maximum day, annual average, and minimum capacity that the facility can operate; each is a factor of the capacity. On this list, the maximum month capacity is listed as a factor of 1, and the others have the factor you would multiply by to get that rating. So the annual average capacity is 80% of themaximum month capacity. When doing your calculations, you might need to run several calculations depending on what you aretrying to determine. Often, the most important calculation is the actual operating rate, which is the measured flow through the facility.

Groundwater wells can present a unique case in that some groundwater pumps are constant speed pumps, so they deliver the amount of water only when they are pumping. Then you have a simpler calculation.

Please note that not all water treatment facilities operate on a 24-hour basis. If a water plant only operates 8 hours per day, then the calculations will need to be adjusted appropriately.


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• Chemical purity and available fluoride ion (AFI)

Chemical Formula Purity Available Fluoride Ion (AFI)

Sodium Fluoride NaF 98% 0.452 Sodium Fluorosilicate Na2SiF6 98.5% 0.607

Fluorosilicic Acid H2SiF6 23% 0.792

The chemical purity of the additive is also important. For the three fluoride products, the available fluoride ion depends on the chemical purity. However, some additives may not meet AWWA specifications and therefore would deliver a different quantity of available fluoride ion than shown in this table. One example is that FSA is nominally sold on a 23% purity basis, but, in fact, each batch might have a different purity. You need to verify thepurity of your batch—is it 22% or 25% pure? And some dry additives can have a lower purity, such as excessive silica in sodium fluoride, which is often the case with imports from othercounties that might not meet the AWWA specification. Make sure you verify the purity of each shipment. The supplier should be able to provide you with a certified test of the batch you receive; you can also run a test of the purity to verify what was delivered.

The available fluoride ion content is based on a 100% purity, soverify the purity of your additive using the AWWA test.


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• Calculated dosage

Rearrange to get . . . .

Fluoride feed rate(lb/day)

=Dosage x Capacity x 8.34

(mg/l) (MGD)

AFI x chemical purity

Dosage(mg/L) =

Feed Rate x AFI X Chemical Purity(lbs/day)

Capacity x 8.34(MGD)

Here is an equation with which you should already be familiar. It’s our old friend, the water formula. As you have learned in other courses, this equation is prepared to allow easy-to-use units, pounds per day, milligrams per liter, and million gallons per day, and the factor 8.34 includes all the conversion factors to make the units resolve. We also include the available fluoride ion and the chemical purity, which we saw on the previous slide.

The first time you run this calculation for your facility, it may take a while to get the calculation correct. But Best Practices for water plant operations are to document all your work in a Standard Operating Procedure, and once you do that, then a daily verification of the calculated feed rate should only take you minute to prepare.If we rearrange this equation to calculate the dosage, we need to multiply both sides of the equation by the AFI and the chemical purity, and divide both sides by the capacity and the 8.34 conversion factor. This results in the lower equation.


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• Quantity to treat 1 MG at 1 mg/L−Sodium fluoride (98%) - 18.8 lbs−Sodium fluorosilicate (98%) - 14.0 lbs−Fluorosilicic acid (23%) -

45.7 lbs. ~ 4.6 gal

With this equation, which is a familiar friend, we can calculatehow much additive is required to treat the flow. As a comparison, to treat 1 million gallons per day with 1 mg of fluoride per liter, presuming the water does not contain any naturally occurring fluoride, the following amount of additive is required:

Sodium fluoride (98% verify the purity, particularly if it is imported), 18.8 lb

Sodium fluorosilicate (98%), 14.0 lb

FSA (23%, and this can vary from delivery to delivery), 45.7 lb/≈4.6 gallons.


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Calculation Problem 1

• Plant 1 has an average daily flow of 1.4 MGD and the source water has a natural fluoride level of 0.15 mg/L. The optimal level for oral health is 0.9 mg/L. If the FSA has a concentration of 25%, what is the dosage required and how many gallons will be necessary?

You will have to compute the usage of fluoride additive at your own facility for the flows and concentrations you will be dealing with. Let’s take a little time to do a few sample calculations. We will do one calculation each for FSA, sodium fluorosilicate, andsodium fluoride, but even though you need to deal with only one of these at your plant, the practice of making the calculation for each example will help you in your management of fluoride levelsin your plant.

For the first example, we have a plant that is treating 1.4 MGD.The natural fluoride in the source water is 0.15 mg/L and the optimal level for oral health for the state is 0.9 mg/L. The FSA has been evaluated by using the hydrogen titration method from the AWWA standard for FSA and has been determined to be 25%. Let’s discuss how to calculate dosage and gallons of solution to be fed.


35

Calculation Problem 1 (step 1)

Average daily flow is 1.4 MGDOptimal fluoride level is 0.9 mg/lNatural fluoride level is 0.15 mg/l

Optimal minus natural is the dosage 0.9 – 0.15 = 0.75 mg/L

We have a plant that is treating 1.4 MGD. The natural fluoride in the source water is 0.15 mg/L and the optimal level for oral health for the state is 0.9 mg/L. So 0.9 minus 0.15 is 0.75 mg/L of fluoride to be added.

Now that we know how much fluoride product to add, we can calculate the number of gallons we need to add each day.


36


Average daily flow of 1.4 MGDOptimal minus natural is

0.9 – 0.15 = 0.75 mg/L

AFI PurityFeed Rate

Fluoride feed rate(lb/day) =

Dosage x Capacity x 8.34(mg/l) (MGD)


44 (lb/day)0.75 (mg/l)

0.792 x 0.25= x 1.4 (MGD) x 8.34

So we calculated that we needed a dosage of 0.9 minus 0.15 which is 0.75 mg/L of fluoride to be added. We put that dosage into the equation, and then substitute the capacity with 1.4 MGD. The FSA has a purity of 25%, which means it has 25% acid and 75% water. And from the earlier table, we know 0.792 mg of fluoride ion is available for each mg of pure acid (assuming 100% purity). Using the standard water treatment equation and substituting the known values in the equation, we can compute that we need to feed 44.22 lb of solution per day. From other math courses, you have probably learned the concept of significant figures in calculations; because the 1.4 MGD has only two significant figures, we can rely on only two significant figures in the final calculations, so we round it to 44 lb/day.

So that is how much fluoride additive is needed in one day.


37



0.9 – 0.15 = 0.75 mg/L

44 lbs/day divided by 24 hours is 1.8 lbs per hour44 lbs/day divided by 24 hours is 1.8 lbs per hour

FSA at 25 percent purity weighs 10.1 pounds per gallon to give FSA at 25 percent purity weighs 10.1 pounds per gallon to give 0.18 pounds per hour0.18 pounds per hour

The total feed rate isThe total feed rate is

4.4 gallons per day (44 lbs divided by 10.1 pounds per gallo4.4 gallons per day (44 lbs divided by 10.1 pounds per gallon)n)

0.18 gallons per hour (4.4 gal per day divided by 24 hours)0.18 gallons per hour (4.4 gal per day divided by 24 hours)

690 mL per hour (4.4 gal multiplied by 3.78 liters per gall690 mL per hour (4.4 gal multiplied by 3.78 liters per gallon)on)

44 (lb/day) = 0.75 (mg/l) x 1.4 (MGD) x 8.340.792 x 0.25

So we have calculated that we need 44 lb/day.

Pounds of solution may be difficult to use as a control amount because the pump is calibrated in gallons, so let’s divide 44 by 10.1, which is how much each gallon of fluorosilicic acid at 25% purity weighs. This gives us 4.4 gallons, which is equal to 0.18gallon per hour (dividing by 24 hours in a day). This is also 690 mL per hour. That is how much we need to set our pump to deliver.

If the water facility does not operate 24-hours per day, then the calculation would need to be adjusted accordingly. If there is only an 8-hour operation, you would divide 4.4 gallons per day by 8 hours to get 0.55 gallons per hour.


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• Plant 2 has an average daily flow of 5.8 MGD and the source water has a natural fluoride level of 0.2 mg/L. The optimal level for oral health is 0.8 mg/L. The fluoride product is sodium fluorosilicate with 98% purity. What is the dosage required and how many pounds will be necessary?

For our second example, we have a plant that is treating 5.8 MGD. The natural fluoride in the source water is 0.2 mg/L, and the optimal level for oral health for the state is 0.8 mg/L. Thesodium fluorosilicate purity has been evaluated by the specific ion electrode method from the AWWA standard and has been determined to be 98%. So let’s discuss how to calculate dosage and pounds to be fed.


39


Average daily flow is 5.8 MGD

Optimal fluoride level is 0.8 mg/L

Natural fluoride level is 0.2 mg/L

Optimal minus natural is 0.8 – 0.2 = 0.6 mg/L

We have a plant that is treating 5.8 MGD. The natural fluoride in the source water is 0.2 mg/L and the optimal level for oral health for the state is 0.8 mg/L. So 0.8 minus 0.2 results in a dosage of 0.6 mg/L of fluoride to be added. Now, we can calculate the amount of additive to be added based on the dosage.


40



0.8 – 0.2 = 0.6 mg/L

AFI PurityFeed Rate




49 (lb/day)0.6 (mg/l)

0.607 x 0.98= x 5.8 (MGD) x 8.34

So we calculated that we needed a dosage of 0.8 minus 0.2 which is 0.6 mg/L of fluoride to be added. We put that dosage into the equation, and then substitute the capacity with 5.8 MGD. The NFSA has a purity of 98%. And from the earlier table, we know 0.607 mg of fluoride ion is available for each mg of additive. Using the standard water treatment equation and substituting the knownvalues in the equation, we can compute that we need to feed 49 lb of solution per day. From other math courses, you have probably learned the concept of significant figures in calculations; because the 5.8 MGD has only two significant figures, we can rely on only two significant figures in the final calculations, so we round it to 49 lb/day.



41


Average daily flow of 5.8 MGDOptimal minus natural is 0.8 – 0.2 = 0.6

mg/L

The feed rate is 49 pounds per day, or The feed rate is 49 pounds per day, or

2.0 pounds per hour (dividing by 24 hours)2.0 pounds per hour (dividing by 24 hours)

0.92 kilograms per hour (divide pounds by 2.2 to get Kg)0.92 kilograms per hour (divide pounds by 2.2 to get Kg)

15 mg per minute (1000 mg in a Kg, and 60 minutes in an hour)15 mg per minute (1000 mg in a Kg, and 60 minutes in an hour)

49 (lb/day) =Dosage x Capacity x 8.34

(mg/l) (MGD)

0.607 x 0.98

So we need to add 49 lb/day. That is how much we need to set our dry feeder to deliver. You can set it based on a calibration test to deliver the precise amount.

If the water facility does not operate 24-hours per day, then the calculation would need to be adjusted accordingly. If there is only an 8-hour operation, you would divide 49 pounds per day by 8 hours to get 6.1 pounds per hour.


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• Plant 3 has an average daily flow of 0.45 MGD and the ground water has a natural fluoride level of 0.4 mg/L.

• The optimal level for oral health is 1.1 mg/L. The fluoride product is sodium fluoride with 96% purity.

What is the dosage required and how many pounds will be necessary?

For our last example, we have a plant that is treating 0.45 MGD.The natural fluoride in the source water is 0.4 mg/L and the optimal level for oral health for the state is 1.1 mg/L. The sodium fluoride purity has been evaluated by the specific ion electrodemethod from the AWWA standard and has been determined to be 96%. Let’s discuss how to calculate dosage and pounds to be fed.


43


Average daily flow is 0.45 MGDOptimal fluoride level is 1.1 mg/LNatural fluoride level is 0.4 mg/LOptimal minus natural is

1.1 – 0.4 = 0.7 mg/L

We have a plant that is treating 0.45 MGD. The natural fluoride in the source water is 0.4 mg/L and the optimal level for oral health for the state is 1.1 mg/L. So 1.1 minus 0.4 is 0.7 mg/L of fluoride to be added. So now we know the dosage we want, we can calculate the feed rate.


44



1.1 – 0.4 = 0.7 mg/L

AFI PurityFeed Rate




6.1 (lb/day)0.7 (mg/l)

0.452 x 0.96= x 0.45 (MGD) x 8.34

So we calculated that we needed a dosage of 1.1 minus 0.4 which is 0.7 mg/L of fluoride to be added. We put that dosage into the equation, and then substitute the capacity with 0.45 MGD. The sodium fluoride has a purity of 96%. And from the earlier table, we know 0.452 mg of fluoride ion is available for each mgof dry additive. Using the standard water treatment equation andsubstituting the known values in the equation, we can compute that we need to feed 6.1 lb of solution per day. From other mathcourses, you have probably learned the concept of significant figures in calculations; because the 0.45 MGD has only two significant figures, we can rely on only two significant figures in the final calculations, so we round it to 6.1 lb/day.



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Calculation Problem 3 (step 3)Average daily flow of 0.45 MGDOptimal minus natural is

1.1 – 0.4 = 0.7 mg/L

The saturator would need an additional 6.1 pounds per day. If The saturator would need an additional 6.1 pounds per day. If a dry feeder is used, then 6.1 pounds divided by 24 hours is a dry feeder is used, then 6.1 pounds divided by 24 hours is 0.25 pounds per hour (or 4 ounces or 120 grams).0.25 pounds per hour (or 4 ounces or 120 grams).

6.1 (lb/day) =0.7 (mg/l) x 0.45 (MGD) x 8.34

0.452 x 0.96

So we need to add 6.1 lb/day. That is how much we need to set our dry feeder to deliver, or the amount that would need to be added to a saturator each day. You would probably not add this much to a saturator each day, but if you don’t maintain the additive bed, the material can be depleted in several days. You can set the dry feeder based on a calibration test to deliver the precise amount.

If the water facility does not operate 24-hours per day, then the calculation would need to be adjusted accordingly. If there is only an 8-hour operation, you would divide 6.1 pounds per day by 8 hours to get 0.76 pounds per hour.


46

Calculation Problem 3 (step 4)Average daily flow of 0.45 MGDOptimal minus natural is

1.1 – 0.4 = 0.7 mg/L

The saturator would need an additional 6.1 pounds per day. The saturator would need an additional 6.1 pounds per day.

A saturated solution of sodium fluoride is 40,000 mg/L, which A saturated solution of sodium fluoride is 40,000 mg/L, which is 18,000 mg/L of fluoride ion. is 18,000 mg/L of fluoride ion.

6.1 pounds per day is 2,770 grams per day, or 2,770,000 mg6.1 pounds per day is 2,770 grams per day, or 2,770,000 mg

2,700,000 mg, divided by 40,000 mg/l is 69 liters per day2,700,000 mg, divided by 40,000 mg/l is 69 liters per day

69 liters per day is 2.9 liter per hour or 18 gallons per day69 liters per day is 2.9 liter per hour or 18 gallons per day

18 gallons per day is 0.76 gallons per hour (48 18 gallons per day is 0.76 gallons per hour (48 mLmL/min) /min)

6.1 (lb/day) =0.7 (mg/l) x 0.45 (MGD) x 8.34

0.452 x 0.96

With this being a saturator, we need to know how much solution to pump in addition to how much sodium fluoride to add to the saturator. 6.1 pounds per day can be multiplied by 454 to get grams per day, and with 1,000 mg per gram, we need 2,770,000 mg of sodium fluoride each day. Since the solution is 40,000 mg/L, if we divide by 40,000, we get 69 liters per day, or 2.9 liters each hour. This is also equal to 0.76 gallons per hour, or 48.3mL/min depending on the pump rating.


47

Simple Rule for Saturated Sodium fluoride solutions

• Based on a ratio of the desired dosage to saturated solution

Fluoride Feed (GPM) =Capacity (GPM) x dosage (mg/l)

18,000 mg/l__________________________

0.0121 GPM =312 (GPM) x 0.7 (dosage)

18,000 (mg/l)_________________________

Compare to previous example: 0.45 MGD @ 0.7 mg/L

0.45 MGD is 312 GPM

0.73 GPH versus 0.76 GPH is within 4 percent

Or 0.73 GPH

( )

This is known as “Robert’s Rule. Robert was Robert Schieferstein who was a founder of LMI pumps.

There is a simple dosage calculation for saturated sodium fluoride solutions. The simple concept here is based on the fact that a saturated sodium fluoride solution will have a definite concentration of fluoride. So we use 1 gallon of saturated solution to treat 18,000 gallons of water to 1ppm. Remember that a saturated solution is 40,000 mg/L of NaF, but 18,000 mg/L of F.

If you want to know how much water you can treat to 1.2ppm its 18,000 / 1.2 = 15,000If you want to know how much water you can treat to .8ppm its 18,000 / .8 = 22,500

For the previous example, the result is 0.73 GPH which is within 4 percent of the previous answer of 0.76 GPH. The 0.76 GPH answer is less accurate for it has several additional rounding errors through all the conversions (0.73 only had one rounding error, while 0.76 had 4 rounding errors). However, the more detailed calculation does give us additional information such ashow much sodium fluoride make-up should be added.


48

Pump Sizing

• Pumps (and dry feeders) operate best in the middle range, not the extreme ranges

• Select pump that will operate predominately in middle range, but is capable at high demand to provide peak delivery: avoid excessive feed

• 10-States specify that the feed rate of solution cannot be less than 20 percent of the maximum pump rated capacity

• Configure pump for proper feed range− Diaphragm pump can provide feed range 1:500

− Peristaltic pump can provide feed range 1:200

Pumps and dry feeders operate best in the middle of their capacity range, not at the extreme ends of their ranges.

Select a pump that will operate predominately in the middle of its range, but is capable of high demand at peak delivery. 10-States require that the feed rate of the solution cannot be less than 20 percent of the maximum rated pump capacity.

Most of the chemical supply pumps can achieve a remarkable delivery range, diaphragm pumps can provide a feed from 1 up to 500, and peristaltic pumps can provide feed ranges from 1 up to 200. This is done by varying diaphragms or tubing sizes, pump stokes or rotation speed, and other techniques. Make sure that the pump is configured to operate in the range that you need foryour facility, not necessarily at its maximum delivery.

Dry feeders can also achieve a wide range of delivery by changing the auger or speed. Ensure that the right equipment is installed.


49

Diaphragm Pump Selection

Example of 1 MGD dosage of 0.8 mg/L

Average feed at rated capacity 3.3 GPDPump delivery for rated capacity 0.14 GPHPeak day factor (2X average) 0.28 GPHMinimum delivery factor 0.6 0.08 GPH*

Use LMI Series P13

*10-States requires minimum delivery to be no less than 20% of maximum rated capacity of pump

For purposes of instruction, let us look at a 1 MGD capacity with 0.8 mg/L dosage requiring 3.3 gallons per day, or 0.14 gph on average at design flow. Presuming maximum day rate of 2 times average design flow, and minimum rate of 60% of the average design flow to allow for lower plant flows in the early years of the system before it reaches the design demand basis, the pump would need to deliver 0.08 to 0.28 gph range. LMI Series P13 has range of 0.001 to 0.42 gph using minimum stroke of 0.6 and maximum stroke of 60, and minimum stroke length of 30 percent.


50

LMI P-13

Select maximum delivery, 0.42 GPH exceeds 0.28 GPH capacity

Verify minimum delivery range 0.0042

GPH with 20% of rated capacity 0.8 GPH

CDC does not endorse particular manufacturers

Average feed.......... 3.3 gpdRated capacity…….. 0.14 gphPeak day…………… 0.28 gphMinimum…………….. 0.08 gph

Checking the manufacturer’s specification sheet, we verify that this pump model can deliver the maximum delivery of 0.42, and also meet the minimum delivery range. The true minimum delivery is actually less than this for the minimum of 0.0042 GPH is based on the number of strokes per minute, and the stroke length of the pump can be further adjusted to 30 percent of the length, so the pump can probably be adjusted to delivery only 0.0013 GPH.


51

Peristaltic Pump Selection

Example of 1.5 MGD dosage of 0.8 mg/L

Average feed at rated capacity 5.0 GPDPump delivery for rated capacity 13 mL/minPeak day factor (2X average) 26 mL/minMinimum delivery day 0.6 X average 5.1 mL/min

Use Watson Marlow Model 520RC (6 – 63 RPM) Using 1.6 mm Tygon tubing

For purposes of identifying a pump, let us look a 1 MGD capacitywith 0.8 mg/L dosage requiring 3.3 gallons per day, or 0.14 gphon average at design flow. Presuming maximum day would be 2 times average for peak day, and minimum rate of 60% of the average rate for the early years when the plant was not at the future design capacity, the pump would need to deliver 3.4 to 19mL/min range. Watson-Marlow has a model 520RC which delivers this range using a 1.6 mm Tygon tubing.


52

Watson Marlow Model 520RC

Select delivery range 2.8 to 28 mL/min

CDC does not endorse particular manufacturers

Design capacity 13 mL/min

Peak day 26 mL/min

Minimum delivery 5.0 mL/min

10-States requires minimum 20% of max rated capacity

5.0 vs 5.6 mL/minis 18%

In this example, we have chosen a pump with a delivery pressure of up to 30 psi, having the delivery range from 2.8 to 28 mL/min to cover our range of 5.0 to 26 mL/min. We selected a tubing diameter to limit the delivery so that we will not over-dose the system. You can further increase or decrease the delivery of this pump by changing the tubing diameter.

10-States requires a minimum of 20% of maximum rated capacity, and in this case, we are slightly below. But it is 18% of maximum capacity, and would probably be considered reasonable.

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